restriction of a function $f$ to the domain $(-\infty,0)$

Question

Consider the function $$f:\mathbb{R}\setminus\{0\}\to\mathbb{R} \ : \ f(x)= {x^3+9\over x^2}$$ And $g$ the limitation of $f$ to the domain $(-\infty,0)$. Prove that $g$ is 1-1 and find the values of the function $g$.

I do not now where to start. Any help is appreciated.

Answer 1

A way to approach such questions for a smooth functions is is to check whether the derivative of the function is positive (or negative) for all values in the domain, and then check the limits at the end points of the domain. Namely $$ \frac{d}{dx}g(x)= \frac{d}{dx}(x+9x^{-2})= 1-18x^{-1}>0,\quad x\in(-\infty,0), $$ which implies that the function is 1-1. Now check the limits $$ \lim_{x\to-\infty} g(x)=-\infty,\quad\text{and}\quad\lim_{x\uparrow 0} g(x)=\infty, $$ which proves that $g((-\infty,0))=\mathbb R$. Definitely useful to make sure you can prove why this is enough.

Answer 2

To prove a function $g(x)$ is one-to-one it means that $ g(x) = g(y)$ implies that $x = y$. For your case you must prove that $$ \frac{x^3 +9 }{x^2} = \frac{y^3 +9 }{y^2} $$ implies that $$x = y$$ When $x$ and $y$ are less than zero.